Results 1 to 4 of 4

- Jan 13th 2009, 06:17 AM #1

- Joined
- Nov 2007
- Posts
- 329

- Jan 13th 2009, 06:35 AM #2

- Jan 13th 2009, 07:30 AM #3

- Joined
- Nov 2007
- Posts
- 329

- Jan 13th 2009, 07:56 AM #4

- Joined
- Nov 2008
- From
- France
- Posts
- 1,458

Hi

Let f(x) = tan x - x - x^3/3

Then f'(x) = tan²x - x² = (tan x -x)(tan x + x)

For 0 < x < pi/2 tan x > 0 and x > 0 therefore tan x + x > 0

The sign of f'(x) is the same as g(x) = tan x - x

g'(x) = tan²x > 0

g is increasing on [0,pi/2[

g(0) = 0 so g(x) > 0 on [0,pi/2[

Therefore f'(x) > 0 on [0,pi/2[

f(0) = 0

f(x) > 0 on [0,pi/2[